Python scoping in functions

Question

Consider the following code:

>>> b = 3
>>> def change (b):
...     b = 2
... 
>>> change(b)
>>> print(b)
3

I am not confused by the fact that the print statement returns 3. Here's the question: is this because of scope, or because the argument that the function takes is not related to the outside b?

So is the b being used in the body of the function the same as the b I defined as 3? — goodcow, Apr 16 '14 at 18:07
[This](http://stackoverflow.com/a/370363/2235132) might help. — devnull, Apr 16 '14 at 18:07
It is because the *name* `b` inside is not related to the *name* `b` outside, although the values are related. See previous questions [here](http://stackoverflow.com/questions/11585768/python-how-does-passing-values-work), [here](http://stackoverflow.com/questions/986006/python-how-do-i-pass-a-variable-by-reference), and [here](http://stackoverflow.com/questions/11585768/python-how-does-passing-values-work) (among others). — BrenBarn, Apr 16 '14 at 18:09
@goodcow: no, the `b` in `change` is a local variable. The other `b` is a *global* variable. — Martijn Pieters, Apr 16 '14 at 18:09
possible duplicate of [Python variable scope question](http://stackoverflow.com/questions/370357/python-variable-scope-question) — llrs, May 02 '14 at 14:33

score 1 · Answer 1 · answered Apr 16 '14 at 18:10

The b inside change is of a different scope than b outside of change. Inside the scope of the function, it does not matter what you call the variable before you pass it in - for now, it's going to be called b. As long as you don't return b from change and assign it to your "original" b, the first one won't change.

def change(b):
    b = 2
    return b

change(b)
print(b) # 3
b = change(b)
print(b) # 2

UnX · Answer 2 · 2014-04-16T23:50:12.190

0

It is a scoping problem. b in the function is a local variable. globals() is dictionary which hold all global variables.

b=3

def change (a):
    globals()[a] = 2

print b
# 3
change("b")
print b
# 2

edited Apr 16 '14 at 23:50

answered Apr 16 '14 at 18:09

UnX

421
3
6

1

This doesn't really answer the question that the OP posted. – anon582847382 Apr 16 '14 at 18:11
Even if you `return b`, you don't affect the global `b`. – John Y Apr 16 '14 at 19:13
@John Y, yes, you need to assign the new value to b. – UnX Apr 16 '14 at 23:29

score 0 · Answer 3 · answered Apr 16 '14 at 18:14

0

You are reassigning b. It does not change the parameter. You can test this using the id() function before and after the assignement.

answered Apr 16 '14 at 18:14

njzk2

38,969
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score 0 · Answer 4 · answered Apr 16 '14 at 18:26

The scope answers are partially correct for the object type you are passing.

Check out this link for information about how python "passes by value reference"

http://stupidpythonideas.blogspot.com/2013/11/does-python-pass-by-value-or-by.html

Here is another link that explains pythons Local -> Enclosed -> Global -> Built-in scope resolution with some examples:

http://nbviewer.ipython.org/github/rasbt/python_reference/blob/master/not_so_obvious_python_stuff.ipynb#python_legb

FWIW, the behavior is slightly different if we consider mutable objects:

>>> b = [3]
>>> def change(b):
...    b[0] = 2
...
>>> change(b)
>>> print(b)
[2]
>>>

score 0 · Answer 5 · answered Apr 16 '14 at 19:07

You have asked quite a tricky question, in the sense that the terminology you have used in framing the question is making people give you differing and weird answers.

You've given two choices:

we have a scope issue
b inside the function isn't related to b outside the function

Well, they are both true, and they are really just two ways to say the same thing.

Forgetting about terminology for a moment, realize that there is absolutely no difference between

b = 3
def f(a):
    print(a)
f(b)

and

b = 3
def f(b):
    print(b)
f(b)

The parameter names in a function's signature (the def line of a function) are local to the function, no matter what names are outside the function.

Now, you probably noticed that I sidestepped the issue of assignment within a function. That's because it's harder to explain quickly, but still be understandable and memorable to someone still in the fairly early stages of learning Python (which, no offense, it seems like you are).

Here is some reading that should help you understand fundamental concepts in Python:

score 0 · Answer 6 · answered Apr 16 '14 at 19:14

You need to look at global variables. Just add global b at the beginning of your code and function, and it should work:

>>> global b
>>> b = 3
>>> def change():
...     global b
...     b = 2
... 
>>> b
3
>>> change()
>>> b
2
>>>

You also need to remove the b parameter from def change(b), because that makes b both local and global. You can also return b:

>>> b = 3
>>> def change(b):
...    b = 2
...    return b
...
>>> b
3
>>> b = change(b)
>>> b
2

Python scoping in functions

6 Answers6